Question

calculate \\( \delta h^{\circ} \\) for the reaction shown below given the following data:
\\( 2 \mathrm{na}(s)+2 \mathrm{h}_{2} \mathrm{o}(l) \to 2 \mathrm{naoh}(a q)+\mathrm{h}_{2}(g) \\)

• -312.4 kj
• -571.2 kj
• -282.8 kj
• +211.1 kj
• -853.6 kj

Explanation:

Step1: Recall the formula for calculating $\Delta H^{\circ}$

The formula is $\Delta H^{\circ}=\sum n\Delta H_{f}^{\circ}(\text{products})-\sum m\Delta H_{f}^{\circ}(\text{reactants})$. Here, $n$ and $m$ are the stoichiometric coefficients.
For the reaction $2\text{Na}(s)+2\text{H}_{2}\text{O}(l)\to2\text{NaOH}(aq)+\text{H}_{2}(g)$, $\Delta H_{f}^{\circ}(\text{Na},s) = 0$ (standard state of an element) and $\Delta H_{f}^{\circ}(\text{H}_{2},g)=0$ (standard state of an element).

Step2: Calculate $\sum n\Delta H_{f}^{\circ}(\text{products})$

For products: $n = 2$ (for $\text{NaOH}$) and $n = 1$ (for $\text{H}_{2}$).
$\sum n\Delta H_{f}^{\circ}(\text{products})=2\times(- 426.8)+1\times0=-853.6\ \text{kJ/mol}$

Step3: Calculate $\sum m\Delta H_{f}^{\circ}(\text{reactants})$

For reactants: $m = 2$ (for $\text{H}_{2}\text{O}$) and $m = 2$ (for $\text{Na}$).
$\sum m\Delta H_{f}^{\circ}(\text{reactants})=2\times(-285.5)+2\times0=-571\ \text{kJ/mol}$

Step4: Calculate $\Delta H^{\circ}$

$\Delta H^{\circ}=-853.6-(-571)=- 282.6\ \text{kJ/mol}$

Answer:

-282.6 kJ (the option corresponding to this value in the given multiple - choice list)